First slide

Resistivity of various materials

Question

Two metal wires of identical dimensions are connected in series. If $σ_{1} a n d σ_{2}$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is

Moderate

A

$\frac{σ_{1} + σ_{2}}{σ_{1} σ_{2}}$
B

$\frac{σ_{1} σ_{2}}{σ_{1} + σ_{2}}$
C

$\frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$
D

$\frac{σ_{1} + σ_{2}}{2 σ_{1} σ_{2}}$

Solution

As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be I and A respectively. Then The resistance of the first wire is
$R_{1} = \frac{l}{σ_{1} A} . . . . (i)$
and that of the second wire is
$R_{2} = \frac{l}{σ_{2} A} . . . (i i)$
As they are connected in series, so their effective resistance is
$R_{s} = R_{1} + R_{2}$
$= \frac{l}{σ_{1} A} + \frac{l}{σ_{2} A} (using (i) and (ii))$
$= \frac{l}{A} (\frac{1}{σ_{1}} + \frac{1}{σ_{2}}) . . . (i i i)$
$I f σ_{e f f} i s t h e e f f e c t i v e c o n d u c t i v i t y o f t h e c o m b i n a t i o n, t h e n$
$R_{s} = \frac{2 l}{σ_{eff} A} . . . . (i v)$
Equating eqns. (iii) and (iv), we get
$\frac{2 l}{σ_{eff} A} = \frac{l}{A} (\frac{1}{σ_{1}} + \frac{1}{σ_{2}})$
$\frac{2}{σ_{eff}} = \frac{σ_{2} + σ_{1}}{σ_{1} σ_{2}} \Rightarrow σ_{eff} = \frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

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